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%\setCJKmonofont{Menlo} % 设置等宽中文字体，用于代码块、等宽文本等。设置等宽中文字体为 Menlo
%\setCJKsansfont{PingFang SC} % 设置无衬线中文字体，用于标题、图表标签等。设置无衬线中文字体为 PingFang SC
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%\newcommand{\showsolution}{2} %%设置showsolution=2, 编译生成试卷解答、考完可以发给学生
\newcommand{\showsolution}{3} %%设置showsolution=3, 编译生成试卷解答与评分标准、阅卷与试卷袋归档

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%% 填写课程信息：
\newcommand{\CourseName}{高等代数一}
\newcommand{\CourseNumber}{160360410}
\newcommand{\CourseStudents}{2024 级数学与应用数学、应用统计学}
\newcommand{\CourseTerm}{2024 $\sim$ 2025 学年 第 一 学期}
\newcommand{\ExamAB}{A}
\newcommand{\ExamContents}{本次考试的主要内容是：行列式、线性方程组、矩阵、二次型。
%考试内容覆盖教学大纲的大部分重要内容。
}

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\title{\CourseName 考试 }
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\date{2025年1月}

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\begin{document}

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\begin{center}
\includegraphics [width=0.85\textwidth, height=4.5cm]{lixin-pan-new.eps}
\end{center}

\vspace{-0.5cm}

\begin{center}
{\Large \bf 上海立信会计金融学院期终考试卷  \hspace{0.3cm} \underline{\,\ExamAB\, } 卷 }

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{\,\CourseStudents\,} 《\underline{\,\CourseName\,}》 课程代码：\underline{\,\CourseNumber\,} }

\vspace{0.3cm}

（本场考试属\underline{ \, 闭 \, }卷考试，考试时间共\underline{ \, 90 \,  }分钟，不准使用计算器）共\underline{ \, \pageref{LastPage} \, }页 

\vspace{0.7cm}

班级 \underline{\hspace{3.5cm}} 学号 \underline{\hspace{3.5cm}} 姓名 \underline{\hspace{3.5cm}} 

\end{center}

\vspace{-0.2cm}

\begin{table}[h]
\centering
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|>{\centering\arraybackslash}p{1cm}|>{\centering\arraybackslash}p{1cm}|>{\centering\arraybackslash}p{1cm}|
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|>{\centering\arraybackslash}p{1cm}|>{\centering\arraybackslash}p{1.2cm}|>{\centering\arraybackslash}p{1.2cm}|}
\hline
题号 &一&二&三&四&总分&合成人签名&审核人签名 \\
\hline
得分 $\,\,\,\,\,\,\,\,$ &&&&&&& \\
\hline
\end{tabular}
\end{table}

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\maketitle

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{\Large \bf 上海立信会计金融学院期终考试卷 \,\, \underline{\, \ExamAB \, } 卷\,\, 解答与评分标准}

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{ \, \CourseStudents \, } 《\underline{ \, \CourseName \, }》 课程代码：\underline{ \, \CourseNumber \, }  }

\end{center}

\vspace{0.3cm}

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%
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%
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%% 大题类型与总分
\begin{minipage}[c]{10cm}
{\bf 一、} 选择题（本大题共 5 小题，每小题 3 分，共 15 分）
\end{minipage}


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\begin{enumerate}

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%\newpage 
\item  %选择题第1题
设 $A, B$ 均为 $n$ 阶矩阵，则下列命题正确的是
%
%\ifnum\showsolution=0
\dotfill (\hspace{1cm})
%\fi
%
\begin{tasks}(1) %每行4个选项
\task [A.]  若 $A$ 或 $B$ 不可逆，则 $AB$ 不可逆。 
\task [B.]  若 $A$ 或 $B$ 可逆，则 $AB$ 可逆。
\task [C.]  若 $A$ 和 $B$ 均不可逆，则 $A+B$ 不可逆。
\task [D.]  若 $A$ 和 $B$ 均可逆，则 $A+B$ 可逆。
\end{tasks}

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%%选择题第1题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：A. 

\begin{tasks}(1) %每行2个选项
\task [A.]  若 $AB$ 可逆，则存在 $C$ 使得 $(AB)C=E$ 以及 $C(AB)=E$. 根据矩阵乘法的结合律，可得 $A(BC)=E$ 以及 $(CA)B=E$. 这样就得出 $A,B$ 都可逆的结论。
\task [B.]  只有在 $A$ 和 $B$ 都可逆的时候，乘积矩阵 $AB$ 才可逆。
\task [C.]  反例：$A=\begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}, B=\begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}$. 
\task [D.]  反例：$A=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}, B=\begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}$. 
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第2题
设 $A$ 为 $m\times n$ 矩阵，齐次线性方程组 $AX=0$ 只有零解的充要条件是
%
%\ifnum\showsolution=0
\dotfill (\hspace{1cm})
%\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $A$ 的列向量组线性无关。
\task [B.]  $A$ 的列向量组线性相关。
\task [C.]  $A$ 的行向量组线性无关。
\task [D.]  $A$ 的行向量组线性相关。
\end{tasks}

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%%选择题第2题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：A. 
因为 $A$ 是 $m\times n$ 矩阵，所以 $X=(x_1,\cdots,x_n)^T$ 共有 $n$ 个未知数。
因为 $AX=0$ 只有零解，所以 $n-R(A)=0$. 所以 $R(A)=n$. 
\begin{tasks}(1) %每行1个选项
\task [A.]  当 $A$ 的列向量组线性无关时，$R(A)=n$. 
\task [B.]  当 $A$ 的列向量组线性相关时，$R(A)<n$. 
\task [C.]  当 $A$ 的行向量组线性无关时，$R(A)=m$. 得不出 $R(A)=n$. 
\task [D.]  当 $A$ 的行向量组线性相关时，$R(A)<m$. 也得不出 $R(A)=n$. 
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第3题

设 $A, B$ 均为 $n$ 阶方阵，$A\neq O$，且 $|AB|=0$，则
%
%\ifnum\showsolution=0
\dotfill (\hspace{1cm})
%\fi
%
\begin{tasks}(2) %每行2个选项
\task [A.]  $|A|\neq 0$.
\task [B.]  $R(B)<n$.
\task [C.]  齐次线性方程组 $(BA)X=0$ 有非零解。
\task [D.]  $R(A)=n$.
\end{tasks}

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%%选择题第3题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：C. 

\begin{tasks}(1) %每行2个选项
\task [A.] 当 $A\neq O$ 时，仅说明 $A$ 不是零矩阵，非零矩阵的行列式可能还是零，例如 $A=\begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}$. 
\task [B.] 反例：$A=\begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix}$, $B=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}$. 这时仍有 $|AB|=0$, 但 $R(B)=n$. 
\task [C.]  因为 $|AB|=0$, 根据行列式乘积公式，有 $|A||B|=0$, 所以 $|B||A|=0$, 又根据行列式乘积公式，有 $|BA|=0$, 所以 $R(BA)<n$, 所以齐次线性方程组 $(BA)X=0$ 有非零解。
\task [D.]  $R(A)=n$ 与 $R(A)<n$ 都是有可能的。
\end{tasks}


}

\fi
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\ifnum\showsolution=0 \newpage \fi  
\item  %选择题第4题
设 $A, B$ 均为 $m\times n$ 矩阵，不能判定 $A$ 和 $B$ 等价的是
%
%\ifnum\showsolution=0
\dotfill (\hspace{1cm})
%\fi
%
\begin{tasks}(1) %每行2个选项
\task [A.]  $R(A)=R(B)$.
\task [B.]  存在可逆矩阵 $P$ 和 $Q$，使得 $A=PBQ$.
\task [C.]  $A$ 和 $B$ 的列向量组可以相互线性表示。
\task [D.]  $A$ 可经初等变换变成 $B$.
\end{tasks}

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%%选择题第4题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：A. 
矩阵 $A$ 与 $B$ 称为等价，是指矩阵 $A$ 可以通过一系列的初等变换得到矩阵 $B$. 

\begin{tasks}(1) %每行2个选项
\task [A.]  若在同一个数域内，则矩阵 $A$ 可以通过一系列的初等变换得到矩阵 $B$. 
\task [B.]  对矩阵 $B$ 进行一些初等行变换，相当于左乘一个可逆矩阵 $P$ 得到 $PB$; 然后再进行一些初等列变换，相当于右乘一个可逆矩阵 $Q$, 得到 $(PB)Q$. 
\task [C.]  先把两个矩阵进行转置。如果 $A$ 和 $B$ 的行向量组可以相互线性表示，那么它们的行最简形是相同的。因此矩阵 $A$ 可以通过一些初等行变换得到矩阵 $B$. 
\task [D.]  这是两个矩阵等价的定义。
\end{tasks}

}

\fi

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%\newpage 
\item  %选择题第5题
设 $A=\left(
  \begin{array}{ccc}
    1 & 1 & 0 \\
    1 & k & 0 \\
    0 & 0 & k-2 \\
  \end{array}
\right)$ 为正定矩阵，$k$ 的取值为
%
%\ifnum\showsolution=0
\dotfill (\hspace{1cm})
%\fi
%
\begin{tasks}(4) %每行2个选项
\task [A.]  $k<1$
\task [B.]  $1<k<2$
\task [C.]  $k>2$
\task [D.]  $1\leq k\leq 2$
\end{tasks}
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%%选择题第5题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：C. 
根据正定矩阵的判别方法，所有的顺序主子式都大于零。因此 
$$
A_1=1>0,\,\, 
A_2=\begin{vmatrix} 1&1 \\ 1&k \end{vmatrix}>0,\,\,  
A_3=\begin{vmatrix} 1&1&0 \\ 1&k&0 \\ 0&0&k-2 \end{vmatrix}>0. $$ 
由此可得 $k>2$. 

}

\fi
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%选择题结束
\end{enumerate}



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%% 试卷：得分框
\ifnum\showsolution=0
%
\hspace{-1.5cm}{\renewcommand{\arraystretch}{2}
\begin{tabular}{|>{\centering\arraybackslash}p{0.8cm}|>{\centering\arraybackslash}p{1.2cm}|} \hline 得分 &  \\  \hline \end{tabular}}
\hspace{0.3cm}
%
\fi
%
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%% 大题类型与总分
\begin{minipage}[c]{10cm}
{\bf 二、} 填空题（本大题共 5 小题，每小题 3 分，共 15 分）
\end{minipage}


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\begin{enumerate}\itemsep0.2cm

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%\newpage 
\item  %填空题第1题
二次型 $f(x,y,z)=-x^2-y^2-z^2+xy+xz+yz$ 的矩阵是 \underline{\hspace{4cm}}。

\vspace{0.2cm}

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%%填空题第1题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：
这个二次型的矩阵是
$$\left(
  \begin{array}{ccc}
    -1 & \frac{1}{2} & \frac{1}{2} \\
   \frac{1}{2} & -1 &\frac{1}{2} \\
    \frac{1}{2} & \frac{1}{2} & -1 \\
  \end{array}
\right). $$ 
注意矩阵不能写成行列式的样子。

}


\fi

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%\newpage 
\item  %填空题第2题

$A$ 为 $3$ 阶方阵，$|A|=\frac{\displaystyle 1}{\displaystyle 2}$，则 $|(2A)^{-1}-5A^*|=$ \underline{\hspace{4cm}}。

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%%填空题第2题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：
根据 ${A}^{-1} = \frac{1}{|{A}|}{A}^*$ 可得 
${A}^* = |{A}| {A}^{-1}$. 因此 
\begin{equation*}
\begin{aligned}
\left\lvert (2{A})^{-1}-5{A}^* \right\rvert 
&= \left\lvert \frac{1}{2}{A}^{-1}-5|{A}|{A}^{-1} \right\rvert 
= \left\lvert \frac{1}{2}{A}^{-1}-\frac{5}{2}{A}^{-1} \right\rvert \\ 
&= \left\lvert -2{A}^{-1} \right\rvert 
= \left(-2\right)^3 \left\lvert{A}^{-1} \right\rvert 
= -8\cdot 2 = -16. 
\end{aligned}
\end{equation*}

}


\fi

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item  %填空题第3题
已知 $A=\left(
  \begin{array}{ccc}
    1 & 1 & 1  \\
   1 & 2 &3 \\
    1 & 4 &9 \\
  \end{array}
\right)$，那么 $M_{21}+M_{22}+M_{23}$=\underline{\hspace{4cm}}。
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%填空题第3题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：
根据余子式和代数余子式的定义，以及行列式按第二行展开的计算方法，
$$
M_{21}+M_{22}+M_{23}
=
-A_{21}+A_{22}-A_{23}
=
\begin{vmatrix}
1&1&1 \\ 
-1&1&-1 \\ 
1&4&9 \\ 
\end{vmatrix}
=16. 
$$
也可以按定义直接计算3个余子式的值。
}


\fi

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%\newpage 
%\item  %填空题第4题
%已知向量组 $\alpha_1=(1,0,1)^T, \alpha_2=(2,2,3)^T, \alpha_3=(1,3,t)^T$ 线性无关，则 $t$ 的取值为 \underline{\hspace{4cm}}。
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%填空题第4题：解答与评分标准
%\ifnum\showsolution>0

%{\color{red} 解答：$t\neq \frac{\displaystyle 5}{\displaystyle 2}$}


%\fi

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%\newpage 
\item  %填空题第5题
齐次线性方程组 $x_1+x_2+\cdots+x_n=0$，则它的基础解系中所含\underline{\hspace{2cm}} 个向量，基础解系为\underline{\hspace{4cm}}（具体向量）。

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%%填空题第5题：解答与评分标准
\ifnum\showsolution>1

{\color{red} 解答：
系数矩阵是 $A=(1,1,\cdots,1)$. 因此基础解系中包含的向量个数为 $n-R(A)=n-1$ 个。
一个基础解系可以取为 $(-1,1,0,\cdots,0)^T, \cdots, (-1,0,0,\cdots,1)^T. $

}

\fi

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%填空题结束
\end{enumerate}

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%\newpage
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%% 试卷：得分框
\ifnum\showsolution=0
%
\hspace{-1.5cm}{\renewcommand{\arraystretch}{2}
\begin{tabular}{|>{\centering\arraybackslash}p{0.8cm}|>{\centering\arraybackslash}p{1.2cm}|} \hline 得分 &  \\  \hline \end{tabular}}
\hspace{0.3cm}
%
\fi
%
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%% 大题类型与总分
\begin{minipage}[c]{10cm}
{\bf 三、} 计算题（本大题共 4 小题，每小题 15 分，共 60 分）
\end{minipage}


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\begin{enumerate}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %计算题第1题
已知 $A, B$ 均为三阶方阵, 将 $A$ 中的第 2 行的 $(-3)$ 倍加至第 3 行得到矩阵 $A_1$, 将 $B$ 中的第 1 列加至第 2 列得到矩阵 $B_1$. 已知$A_1B_1=\begin{pmatrix} 1&4&1 \\ 9&0&4 \\ 0&1&0 \end{pmatrix}$, 求 $AB$. 

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\ifnum\showsolution=0
\vspace{6cm}
\fi

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%%计算题第1题：解答与评分标准
\ifnum\showsolution>1

{\color{red}解答：根据初等行变换相当于左乘初等矩阵，初等列变换相当于右乘初等矩阵，可得 
\begin{eqnarray*} A_1 B_1&=&\left(
         \begin{array}{ccc}
           1 & 0 & 0 \\
           0 & 1 & 0 \\
           0 & -3 & 1 \\
         \end{array}
       \right)A B\left(
         \begin{array}{ccc}
           1 & 1 & 0 \\
           0 & 1 & 0 \\
           0 & 0 & 1 \\
         \end{array}
       \right). 
\end{eqnarray*}
%
\ifnum\showsolution=3
\dotfill (\underline{\,\,6\text{分}\,\,})
\fi
%

将 $AB$ 左右两边的矩阵移项可得 
\begin{eqnarray*} 
AB&=&\left(
         \begin{array}{ccc}
           1 & 0 & 0 \\
           0 & 1 & 0 \\
           0 & 3 & 1 \\
         \end{array}
       \right)A_1B_1\left(
         \begin{array}{ccc}
           1 & -1 & 0 \\
           0 & 1 & 0 \\
           0 & 0 & 1 \\
         \end{array}
       \right)
=\left(
         \begin{array}{ccc}
           1 & 3 & 1 \\
           9 & -9 & 4 \\
           27 & -26 & 12 \\
         \end{array}
       \right).
 \end{eqnarray*} 
%
\ifnum\showsolution=3
\dotfill (\underline{\,\,第1个等号 6分，第2个等号3分\,\,})
\fi
%

}

\fi


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\ifnum\showsolution=0 \newpage \fi  
\item %计算题第2题
设有三维向量 $\alpha_1=(1+t, 1, 1)^T, \alpha_2=(1, 1+t, 1)^T, \alpha_3=(1, 1, 1+t)^T, \beta=(0,t, t^2)^T$. 
问 $t$ 取何值时,
\begin{enumerate}[label={(\arabic*)}]
\item $\beta$ 可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示, 且表达式唯一；
\item $\beta$ 可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示, 且表达式不唯一；
\item $\beta$ 不可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示。
\end{enumerate} 

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{\color{red}解答一：
向量 $\beta$ 可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示，是指存在实数 $x_1,x_2,x_3$ 使得 
$\beta= x_1\alpha_1+ x_2\alpha_2+x_3\alpha_3$. \\ 
具体写出来，就是向量方程
\begin{equation*}
x_1\begin{pmatrix}  1+t \\ 1 \\ 1  \end{pmatrix} 
+ x_2\begin{pmatrix}  1 \\ 1+t \\ 1  \end{pmatrix} 
+ x_3\begin{pmatrix}  1 \\ 1 \\ 1+t  \end{pmatrix} 
= \begin{pmatrix}  0 \\ t \\ t^2  \end{pmatrix}. 
\end{equation*}
也可以写成矩阵方程
\begin{equation*}
\begin{pmatrix}  1+t & 1 & 1 \\ 1 & 1+t & 1 \\ 1 & 1 & 1+t  \end{pmatrix} 
\begin{pmatrix}  x_1 \\ x_2 \\ x_3  \end{pmatrix} 
= \begin{pmatrix}  0 \\ t \\ t^2  \end{pmatrix}. 
\end{equation*}

系数行列式为 
$$\left|
            \begin{array}{ccc}
              1+t & 1 & 1 \\
              1 & 1+t & 1 \\
              1 & 1 & 1+t\\
            \end{array}
          \right|=t^2(t+3). $$
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\begin{enumerate}[label={(\arabic*)}]
\item  当 $t\neq 0$ 且 $t\neq -3$ 时，方程组有唯一解，这时 $\beta$ 可由 $\alpha_1, \alpha_2, \alpha_3$ 唯一线性表示。
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\item[(2)] 当 $t=0$ 时，方程组有无穷多个解，这时 $\beta$ 可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示且表达式不唯一。
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\item[(3)] 当 $t=-3$ 时，方程组无解，这时 $\beta$ 不可由 $\alpha_1, \alpha_2, \alpha_3$ 线性表示。
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\end{enumerate}

解答二：将上述线性方程组的增广矩阵通过初等行变换化为阶梯形矩阵，可得 
 $$\left(
            \begin{array}{cccc}
              1+t & 1 & 1 & 0\\
              1 & 1+t & 1 & t\\
              1 & 1 & 1+t & t^2\\
            \end{array}
          \right)\rightarrow \left(
            \begin{array}{cccc}
              1 & 1 & 1+t & t^2\\
              0 & t & -t & t(1-t)\\
              0 & 0 & t(t+3) & t(t^2+2t-1)\\
            \end{array}
          \right)$$
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由此得出同样结论。

}
\fi


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%\newpage 
\item %计算题第3题
设 $n$ 阶矩阵 $A, B$ 满足条件 $2A-B-AB=E$, $A^2=A$.

$(1)$ 证明: $A-B$ 为可逆矩阵, 并求 $(A-B)^{-1}$.\\
$(2)$ 已知 $A=\begin{pmatrix}1&0&0\\ 0&3&-1\\ 0&6&-2\end{pmatrix}$, 求矩阵 $B$.

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{\color{red}解答：
\begin{enumerate}[label={(\arabic*)}] 
\item  
根据条件可得 $(E+A)(A-B)=E$.
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因此 $A-B$ 是可逆矩阵，且 $(A-B)^{-1}=E+A$. 
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\item  因为 $A-B=(E+A)^{-1}$, 所以 $B=A-(E+A)^{-1}$. 计算可得 
$$(E+A)^{-1}=\left(
\begin{array}{ccc}
1/2 & 0 & 0\\
0 & -1/2 & 1/2 \\
0 & -3 & 2 \\
\end{array}
\right),
$$
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以及
$$ B=A-(E+A)^{-1}= \left(
\begin{array}{ccc}
1/2 & 0 & 0\\
0 & 7/2 & -3/2 \\
0 & 9 & -4 \\
\end{array}
\right). $$
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%                                                     

\end{enumerate}

}

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\item %计算题第4题
把二次型 $f(x_1, x_2, x_3)=4x_1x_2-2x_1x_3-2x_2x_3+3x_3^2$ 化为标准型，并求相应的非退化线性变换和二次型的符号差。

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{\color{red}解答：

解法一： 
这个二次型的矩阵为 $A=\left(
              \begin{array}{ccc}
                0 & 2 & -1 \\
                2 & 0 & -1 \\
                -1 & -1 & 3 \\
              \end{array}
            \right)
$. 
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对矩阵 $A$ 做合同变换，即依次进行相同的初等行变换和初等列变换，并将初等列变换记录在下方的单位矩阵里，可得 
$$\left(
         \begin{array}{ccc}
           0 & 2 & -1 \\
                2 & 0 & -1 \\
                -1 & -1 & 3 \\
           1 & 0 & 0 \\
           0 & 1 & 0 \\
           0 & 0 & 1 \\
         \end{array}
       \right)\rightarrow \left(
         \begin{array}{ccc}
           1 & 0 & 0 \\
           0 & -1 & 0 \\
           0 & 0 & 8 \\
           1 & -1 & 1 \\
           0 & 1 & -3 \\
           1 & -1 & 2 \\
         \end{array}
       \right). 
$$
因此标准型为 $g(y_1, y_2, y_3)=y_1^2-y_2^2+8y_3^2$. 
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所做的非退化线性变换为 $\left(
\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & -1 & 1 \\
0 & 1 & -3 \\
1 & -1& 2 \\
\end{array}
\right)\left(
\begin{array}{c}
y_1 \\
y_2 \\
y_3 \\
\end{array}
\right)$. 
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二次型 $f$ 的符号差为$2-1=1$.
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解法二：使用配方法。
\begin{equation*}
\begin{aligned}
f(x_1, x_2, x_3) &= 4x_1x_2-2x_1x_3-2x_2x_3+3x_3^2 \\ 
&= 4x_1x_2 + [3x_3^2 -2(x_1+x_2)x_3] \\ 
&= -\frac{1}{3}(x_1-5x_2)^2 + 8x_2^2 + 3(x_3-\frac{1}{3}x_1 - \frac{1}{3}x_2)^2 \\  
& = -\frac{1}{3}y_1^2 + 8y_2^2 +3y_3^2. 
\end{aligned}
\end{equation*}
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所用的非退化的线性变换为 
\begin{equation*}
\left\{
\begin{aligned}
y_1 &= x_1 - 5x_2, \\ 
y_2 &= x_2, \\ 
y_3 &= x_3 - \frac{1}{3}x_1 -\frac{1}{3} x_2, \\ 
\end{aligned}
\right. 
\textrm{ 即 }
\left\{
\begin{aligned}
x_1 &= y_1 + 5y_2, \\ 
x_2 &= y_2, \\ 
x_3 &= \frac{1}{3}y_1 + 2y_2 + y_3. \\ 
\end{aligned}
\right. 
\end{equation*}
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符号差为 $p-q=1$. 
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}
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\end{enumerate}


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\begin{tabular}{|>{\centering\arraybackslash}p{0.8cm}|>{\centering\arraybackslash}p{1.2cm}|} \hline 得分 &  \\  \hline \end{tabular}}
\hspace{0.3cm}
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\begin{minipage}[c]{10cm}
{\bf 四、} 证明题（本大题共 1 小题，共 10 分）
\end{minipage}


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\begin{enumerate}

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%\newpage 
\item %证明题第1题
若向量 $\alpha_1, \cdots, \alpha_s \ (s\geq 2)$ 线性无关，试讨论 $$\beta_1=\alpha_2+\cdots+\alpha_s, \beta_2=\alpha_1+\alpha_3+\cdots+\alpha_s, \cdots, \beta_s=\alpha_1+\alpha_2+\cdots+\alpha_{s-1}$$ 的线性相关性。

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{\color{red}证明一：根据题目条件可得 
$$(\beta_1, \beta_2, \cdots, \beta_s)=(\alpha_1, \alpha_2, \cdots, \alpha_s)\left(
\begin{array}{cccc}
0 & 1 & \cdots & 1 \\
1 & 0 & \cdots & 1 \\
\vdots & \vdots& \ddots & \vdots \\
1 & 1 & \cdots & 0 \\
\end{array}
\right).
$$
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因为上述右边的矩阵的行列式的值不为零，
$$ \left|
\begin{array}{cccc}
0 & 1 & \cdots & 1 \\
1 & 0 & \cdots & 1 \\
\vdots & \vdots& \ddots & \vdots \\
1 & 1 & \cdots & 0 \\
\end{array}
\right|=(-1)^{s-1}(s-1)\neq 0, $$
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%                                                            

所以 $\beta_1, \cdots, \beta_s$ 线性无关。
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证明二：按定义验证。设有线性组合 $k_1\beta_1 + k_2\beta_2 +\cdots + k_s\beta_s =0$. 
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则有 
$$
(\alpha_1, \alpha_2, \cdots, \alpha_s)\left(
\begin{array}{cccc}
0 & 1 & \cdots & 1 \\
1 & 0 & \cdots & 1 \\
\vdots & \vdots& \ddots & \vdots \\
1 & 1 & \cdots & 0 \\
\end{array}
\right)
\begin{pmatrix}k_1 \\ k_2 \\ \vdots \\ k_s  \end{pmatrix} =0.
$$
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因为向量组 $\alpha_1, \alpha_2, \cdots, \alpha_s$ 线性无关，所以有齐次线性方程组 
$$
\left(
\begin{array}{cccc}
0 & 1 & \cdots & 1 \\
1 & 0 & \cdots & 1 \\
\vdots & \vdots& \ddots & \vdots \\
1 & 1 & \cdots & 0 \\
\end{array}
\right)
\begin{pmatrix}k_1 \\ k_2 \\ \vdots \\ k_s  \end{pmatrix} 
=\begin{pmatrix}0 \\ 0 \\ \vdots \\ 0  \end{pmatrix}.
$$
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因为系数矩阵的行列式的值不等于零，所以只有零解
$$
\begin{pmatrix}k_1 \\ k_2 \\ \vdots \\ k_s  \end{pmatrix}
=\begin{pmatrix}0 \\ 0 \\ \vdots \\ 0  \end{pmatrix}.
$$
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}


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\end{enumerate}




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